An inductor of self inductance 1 H is connected in series with a resistor of 100π ohm and an ac supply of 100π volt, 50 Hz. Maximum current flowing in the circuit is _________ A. Numerical answer type. Work it out, then open the answer.

Answer: 1. XL = 2π f L = 2π × 50 × 1 = 100π Ω, and R = 100π Ω. Z = √(R² + XL²) = 100π√(2) Ω Supply voltage is rms: Irms = (100π)/(100π√(2)) = 1/√(2) A Imax = √(2) Irms = 1 A

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RL and RC Series Circuits

An inductor of self inductance 1 H is connected in series with a resistor of $100\pi$ ohm and an ac supply of $100\pi$ volt, 50 Hz. Maximum current flowing in the circuit is _________ A.

Numerical answer type. Work it out, then open the answer.

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Answer: 1

$X_L = 2\pi f L = 2\pi \times 50 \times 1 = 100\pi\ \Omega$ , and $R = 100\pi\ \Omega$ .
$Z = \sqrt{R^2 + X_L^2} = 100\pi\sqrt{2}\ \Omega$
Supply voltage is rms: $I_{rms} = \frac{100\pi}{100\pi\sqrt{2}} = \frac{1}{\sqrt{2}}\text{ A}$
$I_{max} = \sqrt{2}\, I_{rms} = 1\text{ A}$

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